Respuesta :
Answer:
B
Step-by-step explanation:
This is actually in the form of ax+by=c which is called standard form for a line.
I also know it is linear because the degree is 1. I know the degree is 1 because both variables are to the first power (you don't see this number because [tex]x^1=x[/tex] or [tex]y^1=y[/tex]).
So the choices are between A and B.
Let's see if it satisfies A:
Checking (6,2) for (x,y):
3 x - 2 y=4
3(6)-2(2)=4
18 - 4 =4
14 =4 is false so (6,2) is not on the given line.
Moving on to B:
Checking (0,-2) for (x,y):
3 x - 2 y=4
3(0)-2(-2)=4
0 + 4 =4
4 =4 is true so (0,-2) is on the given line.
Checking (2,1) for (x,y):
3 x - 2 y=4
3(2)-2(1)=4
6 - 2 =4
4 =4 is true so (2,1) is on the given line.
Checking (-2,-5) for (x,y):
3 x - 2 y=4
3(-2)-2(-5)=4
-6 + 10 =4
4 =4 is true so (-2,-5) is on the given line.
We have that all three pairs from choice B are contained on the line given.
Answer:
B).
Step-by-step explanation:
3x - 2y = 4 contains variables of degree one only, so this is a straight line.
Convert to slope-intercept form so it will be easier to check:
3x - 2y = 4
-2y = -3x + 4
y = 3/2 x - 2.
A). (6, 2): y = 3/2 *6 - 2 = 9-2 = 7 so its not A.
B). (0, -2): y = 0 - 2 = -2 so this fits.
(2, 1): y = 3/2 * 2 - 2 = 1 so this fits.
(-2, -5): y = 3/2*(-2) - 2 - -3-2 = -5 so this fits.
